Chapter 3: Introduction to SQL

1. Chapter 3: Introduction to SQL

2. Chapter 3: Introduction to SQL • Overview of the SQL Query Language • Data Definition • Basic Query Structure • Additional Basic Operations • Set Operations • Null Values • Aggregate Functions • Nested Subqueries • Modification of the Database

3. History • IBM Sequel language developed as part of System R project at the IBM San Jose Research Laboratory • Renamed Structured Query Language (SQL) • ANSI and ISO standard SQL: • SQL-86, SQL-89, SQL-92 • SQL:1999, SQL:2003, SQL:2008, … • Commercial systems offer most, if not all, SQL-92 features, plus varying feature sets from later standards and special proprietary features. • Not all examples here may work on your particular system.

4. Data Definition Language • The schema for each relation. • The domain of values associated with each attribute. • Integrity constraints • And as we will see later, also other information such as • The set of indices to be maintained for each relations. • Security and authorization information for each relation. • The physical storage structure of each relation on disk. The SQL data-definition language (DDL) allows the specification of information about relations, including:

5. Domain Types in SQL • char(n). Fixed length character string, with user-specified length n. • varchar(n). Variable length character strings, with user-specified maximum length n. • int.Integer (a finite subset of the integers that is machine-dependent). • smallint. Small integer (a machine-dependent subset of the integer domain type). • numeric(p,d). Fixed point number, with user-specified precision of p digits, with n digits to the right of decimal point. • real, double precision, float(n) • More are covered in Chapter 4.

6. Create Table Construct • An SQL relation is defined using thecreate tablecommand: Example: create tableinstructor (IDchar(5),name varchar(20),dept_name varchar(20),salarynumeric(8,2)) • insert into instructor values (‘10211’, ’Smith’, ’Biology’, 66000); • insert into instructor values (‘10211’, null, ’Biology’, 66000);

7. Integrity Constraints in Create Table • not null • primary key (A1, ..., An ) • foreign key (Am, ..., An ) references r Example: Declare branch_name as the primary key for branch . create tableinstructor (IDchar(5),name varchar(20) not null,dept_name varchar(20),salarynumeric(8,2),primary key (ID),foreign key (dept_name) references department) primary key declaration on an attribute automatically ensuresnot null

8. And a Few More Relation Definitions • create tablestudent (IDvarchar(5) primary key,namevarchar(20) not null,dept_namevarchar(20),tot_crednumeric(3,0),foreign key (dept_name) references department) ); • create tabletakes (IDvarchar(5),course_idvarchar(8),sec_idvarchar(8),semestervarchar(6),yearnumeric(4,0),gradevarchar(2),primary key (ID, course_id, semester, year),foreign key (ID) references student,foreign key (course_id, sec_id, semester, year) references section );

9. And more still • create tablecourse (course_idvarchar(8) primary key,titlevarchar(50),dept_namevarchar(20),creditsnumeric(2,0),foreign key (dept_name) references department) ); Quiz Q1: True or false: The primary key constraint ensures that primary key values: (A) are repeated across tuples (B) are not null (1) True, True (2) True, False (3) False, True (4) false, false Quiz Q2: the foreign key constraint on course ensures: (1) all departments have associated courses (2) dept_name is not null, i.e. all courses have departments (3) dept_name occurs in the department relation

10. Drop and Alter Table Constructs • drop table student • Deletes the table and its contents • delete from student • Deletes all contents of table, but retains table • alter table • alter table r add A D • where A is the name of the attribute to be added to relation r and D is the domain of A. • All tuples in the relation are assigned default or null as the value for the new attribute. • alter table r drop A • where A is the name of an attribute of relation r • Dropping of attributes not supported by many databases

11. Basic Query Structure • A typical SQL query has the form:select A1, A2, ..., Anfromr1, r2, ..., rmwhere P • Ai represents an attribute • Ri represents a relation • P is a predicate. • The result of an SQL query is a relation.

12. The select Clause • The select clause list the attributes desired in the result of a query • corresponds to the projection operation of the relational algebra • Example: find the names of all instructors:select namefrom instructor • NOTE: SQL names are case insensitive (i.e., you may use upper- or lower-case letters.) • E.g. Name ≡ NAME ≡ name • Some people use upper case wherever we use bold font.

13. The select Clause (Cont.) • SQL allows duplicates in relations as well as in query results. • To force the elimination of duplicates, insert the keyword distinct after select. • Find the names of all departments with instructor, and remove duplicates select distinct dept_namefrom instructor • The keyword all specifies that duplicates not be removed. select alldept_namefrom instructor

14. The select Clause (Cont.) • An asterisk in the select clause denotes “all attributes” select *from instructor • The select clause can contain arithmetic expressions involving the operation, +, –, , and /, and operating on constants or attributes of tuples. • The query: selectID, name, salary/12from instructor would return a relation that is the same as the instructor relation, except that the value of the attribute salary is divided by 12. Quiz Q3: Which of these clauses is optional in an SQL query: (1) select (2) from (3) where (4) none of these

15. The where Clause • The whereclause specifies conditions that the result must satisfy • Corresponds to the selection predicate of the relational algebra. • To find all instructors in Comp. Sci. dept with salary > 80000 select namefrom instructorwhere dept_name =‘Comp. Sci.'and salary > 80000 • Comparison results can be combined using the logical connectives and, or, and not. • Comparisons can be applied to results of arithmetic expressions.

16. The from Clause • The fromclause lists the relations involved in the query • Corresponds to the Cartesian product operation of the relational algebra. • Find the Cartesian product instructor X teaches select from instructor, teaches • generates every possible instructor – teaches pair, with all attributes from both relations • Cartesian product not very useful directly, but useful combined with where-clause condition (selection operation in relational algebra)

17. Cartesian Product: instructor X teaches instructor teaches

18. Joins • For all instructors who have taught some course, find their names and the course ID of the courses they taught. select name, course_idfrom instructor, teacheswhere instructor.ID = teaches.ID • Find the course ID, semester, year and title of each course offered by the Comp. Sci. department select section.course_id, semester, year, titlefrom section, coursewhere section.course_id = course.course_id anddept_name = ‘Comp. Sci.'

19. Let’s Try Writing Some Queries in SQL Some queries to be written….. • Find the salary of the instructor with ID 10101 • Find titles of all courses in the Comp. Sci. department • Find course ID, year and semester of all courses taken by any student named “Shankar” • As above, but additionally show the title of the course also

20. Quiz Quiz Q4: The query select name, course_idfrom instructor, teaches (1) returns all matching instructor/teaches pairs (2) is a syntax error (3) returns all pairs of instructor, teaches tuples (4) none of the above

21. Natural Join • Natural join matches tuples with the same values for all common attributes, and retains only one copy of each common column • select *from instructor natural join teaches;

22. Natural Join Example • List the names of instructors along with the course ID of the courses that they taught. • select name, course_idfrom instructor, teacheswhere instructor.ID = teaches.ID; • select name, course_idfrom instructor natural join teaches;

23. Natural Join (Cont.) • Danger in natural join: beware of unrelated attributes with same name which get equated incorrectly • List the names of instructors along with the the titles of courses that they teach • Incorrect version (makes course.dept_name = instructor.dept_name) • select name, titlefrom instructor natural join teaches natural join course; • Correct version • select name, titlefrom instructor natural join teaches, coursewhere teaches.course_id= course.course_id; • Another correct version • select name, titlefrom (instructor natural join teaches) join course using(course_id);

24. The Rename Operation • The SQL allows renaming relations and attributes using the as clause: old-name as new-name • E.g. • select ID, name, salary/12 as monthly_salaryfrom instructor • Find the names of all instructors who have a higher salary than some instructor in ‘Comp. Sci’. • select distinct T. namefrom instructor as T, instructor as Swhere T.salary > S.salary and S.dept_name = ‘Comp. Sci.’ • Keyword as is optional and may be omittedinstructor as T ≡ instructorT • Keyword as must be omitted in Oracle

25. String Operations • SQL includes a string-matching operator for comparisons on character strings. The operator “like” uses patterns that are described using two special characters: • percent (%). The % character matches any substring. • underscore (_). The _ character matches any character. • Find the names of all instructors whose name includes the substring “dar”. select namefrom instructorwherename like '%dar%' • Match the string “100 %” like ‘100 \%' escape '\' • SQL supports a variety of string operations such as • concatenation (using “||”) • converting from UPPER to lower case (and vice versa) • finding string length, extracting substrings, etc.

26. Ordering the Display of Tuples • List in alphabetic order the names of all instructors select distinct namefrom instructororder by name • We may specify desc for descending order or asc for ascending order, for each attribute; ascending order is the default. • Example: order bynamedesc • Can sort on multiple attributes • Example: order by dept_name, name Quiz Q5: If you omit the order by clause(1) there is no default ordering(2) the results may turn out to be ordered (3) both the above are true (4) neither 1 nor 2 is true.

27. Duplicates • Multiset: set with duplicates • In relations with duplicates, SQL can define how many copies of tuples appear in the result. • Multisetversions of some of the relational algebra operators – given multiset relations r1 and r2: 1.  (r1): If there are c1 copies of tuple t1 in r1, and t1 satisfies selections ,, then there are c1 copies of t1 in  (r1). 2. A (r ): For each copy of tuple t1in r1, there is a copy of tupleA (t1) in A (r1) where A (t1) denotes the projection of the single tuple t1. 3. r1 x r2 : If there are c1 copies of tuple t1in r1 and c2 copies of tuple t2 in r2, there are c1 x c2 copies of the tuple t1. t2 in r1 x r2

28. Duplicates (Cont.) • Example: Suppose multiset relations r1 (A, B) and r2 (C) are as follows: r1 = {(1, a) (2,a)} r2 = {(2), (3), (3)} • With multiset relational algebraB(r1) would be {(a), (a)}, B(r1) x r2 would be {(a,2), (a,2), (a,3), (a,3), (a,3), (a,3)} • SQL duplicate semantics: select A1,, A2, ..., Anfrom r1, r2, ..., rmwhere P is equivalent to the multiset version of the expression:

29. Set Operations • Find courses that ran in Fall 2009 or in Spring 2010 (selectcourse_id from section where sem = ‘Fall’ and year = 2009)union(selectcourse_id from section where sem = ‘Spring’ and year = 2010) • Find courses that ran in Fall 2009 and in Spring 2010 (selectcourse_id from section where sem = ‘Fall’ and year = 2009)intersect(selectcourse_id from section where sem = ‘Spring’ and year = 2010) • Find courses that ran in Fall 2009 but not in Spring 2010 (selectcourse_id from section where sem = ‘Fall’ and year = 2009)except(selectcourse_id from section where sem = ‘Spring’ and year = 2010)

30. Set Operations • Set operations union, intersect, and except • Each of the above operations automatically eliminates duplicates • To retain all duplicates use the corresponding multiset versions union all, intersect alland except all.Suppose a tuple occurs m times in r and n times in s, then, it occurs: • m + n times in r union all s • min(m,n) times in rintersect all s • max(0, m – n) times in rexcept all s

31. Null Values • It is possible for tuples to have a null value, denoted by null, for some of their attributes • null signifies an unknown value or that a value does not exist. • The result of any arithmetic expression involving null is null • Example: 5 + null returns null • The predicate is null can be used to check for null values. • Example: Find all instructors whose salary is null. select namefrom instructorwhere salary is null • Any comparison with null returns unknown • Returning false could cause problems: • Consider r.A < 10 vs. not (r.A >= 10)

32. Null Values and Three Valued Logic • Any comparison with null returns unknown • Example: 5 < null or null <> null or null = null • Three-valued logic using the truth value unknown: • OR: (unknownortrue) = true, (unknownorfalse) = unknown (unknown or unknown) = unknown • AND: (true and unknown) = unknown, (false and unknown) = false, (unknown and unknown) = unknown • NOT: (not unknown) = unknown • “P is unknown”evaluates to true if predicate P evaluates to unknown • Result of where clause predicate is treated as false if it evaluates to unknown

33. Quiz Quiz Q6: Consider the where clause predicates where r.A < 5 and where not (r.A >= 5) (1) the two are equivalent(2) the two are equivalent only if r.a is declared as not null (3) the two are equivalent only if r.a is null (4) the two are never equivalent

34. Aggregate Functions • These functions operate on the multiset of values of a column of a relation, and return a value avg: average valuemin: minimum valuemax: maximum valuesum: sum of valuescount: number of values

35. Aggregate Functions (Cont.) • Find the average salary of instructors in the Computer Science department • select avg (salary)from instructorwhere dept_name= ’Comp. Sci.’; • Find the total number of instructors who teach a course in the Spring 2010 semester • select count (distinct ID)from teacheswhere semester = ’Spring’ and year = 2010 • Find the number of tuples in the course relation • select count (*)from course;

36. Aggregate Functions – Group By • Find the average salary of instructors in each department • select dept_name, avg (salary)from instructorgroup by dept_name; • Note: departments with no instructor will not appear in result

37. Aggregation (Cont.) • Attributes in select clause outside of aggregate functions must appear in group by list • /* erroneous query */select dept_name, ID, avg (salary)from instructorgroup by dept_name;

38. Aggregate Functions – Having Clause • Find the names and average salaries of all departments whose average salary is greater than 42000 select dept_name, avg (salary) from instructor group by dept_name having avg (salary) > 42000; Note: predicates in the having clause are applied after the formation of groups whereas predicates in the where clause are applied before forming groups

39. Null Values and Aggregates • Total all salaries select sum (salary )from instructor • Above statement ignores null amounts • Result is null if there is no non-null amount • All aggregate operations except count(*) ignore tuples with null values on the aggregated attributes • What if collection has only null values? • count returns 0 • all other aggregates return null Quiz Q7: given a multiset of values {1, 1, 3, 3, null} the average is (1) 1 (2) 2 (3) 8/5 (4) null

40. Let’s Try Writing Some Queries in SQL Some queries to be written….. Find the total number of courses taken by Shankar Find first year in which CS 801 ran. Find IDs and number of courses taken by all students who have taken more than 20 courses

41. Nested Subqueries • SQL provides a mechanism for the nesting of subqueries. • A subquery is a select-from-where expression that is nested within another query. • A common use of subqueries is to perform tests for set membership, set comparisons, and set cardinality.

42. Example Query • Find courses offered in Fall 2009 and in Spring 2010 select distinct course_id from section where semester = ’Fall’ and year= 2009 and course_id in (select course_id from section where semester = ’Spring’ and year= 2010); • Find courses offered in Fall 2009 but not in Spring 2010 select distinct course_id from section where semester = ’Fall’ and year= 2009 and course_id not in (select course_id from section where semester = ’Spring’ and year= 2010);

43. Example Query • Find the total number of (distinct) students who have taken course sections taught by the instructor with ID 10101 select count (distinct ID) from takes where (course_id, sec_id, semester, year) in (select course_id, sec_id, semester, year from teaches where teaches.ID= 10101); • Note: Above query can be written in a much simpler manner. The formulation above is simply to illustrate SQL features. Quiz Q8: The above query can be rewritten using (1) natural join (2) aggregation (3) group by (4) order by

44. Set Comparison • Find names of instructors with salary greater than that of some (at least one) instructor in the Biology department. select distinct T.name from instructor as T, instructor as S where T.salary > S.salary and S.dept name = ’Biology’; • Same query using > some clause select name from instructor where salary > some (select salary from instructor where dept name = ’Biology’);

45. Example Query • Find the names of all instructors whose salary is greater than the salary of all instructors in the Biology department. select name from instructor where salary > all (select salary from instructor where dept name = ’Biology’); Can we write this using aggregates?

46. Example Query Find the names of all instructors whose salary is greater than the salary of all instructors in the Biology department. select name from instructor where salary > all (select salary from instructor where dept name = ’Biology’); Can we write this using aggregates? Yes: select name from instructor where salary > (select max(salary) from instructor where dept name = ’Biology’);

47. Test for Empty Relations • The exists construct returns the value true if the argument subquery is nonempty. • exists r  r  Ø • not exists r  r = Ø

48. Correlation Variables • Yet another way of specifying the query “Find all courses taught in both the Fall 2009 semester and in the Spring 2010 semester” select course_idfrom section as Swhere semester = ’Fall’ and year= 2009 and exists (select *from section as Twhere semester = ’Spring’ and year= 2010 and S.course_id= T.course_id); • Correlated subquery • Correlation name or correlation variable

49. Not Exists • Find all students who have taken all courses offered in the Biology department. select distinct S.ID, S.name from student as S where not exists ( (select course_id from course where dept_name = ’Biology’) except (select T.course_id from takes as T where S.ID = T.ID)); SQA SQB • Note that X – Y = Ø  X Y • Think of above query as select distinct S.ID, S.name • from student as S • where SQB contains SQA • Note: Cannot write this query using= alland its variants

50. Subqueries in From Clause • SQL allows a subquery expression to be used in the from clause • Find the average instructors’ salaries of those departments where the average salary is greater than $42,000. select dept_name, avg_salaryfrom (select dept_name, avg (salary) as avg_salaryfrom instructorgroup by dept_name)where avg_salary > 42000; • Note that we do not need to use the having clause • Another way to write above query select dept_name, avg_salaryfrom (select dept_name, avg (salary) from instructorgroup by dept_name) as dept_avg (dept_name, avg_salary)where avg_salary > 42000;